Are zero sets of polynomial equations closed because of the fundamental theorem of algebra?

Question

Tu Manifolds Section 15.2

Tu Manifolds Section 9.2

Level sets in Tu Manifolds does not look different from fibres in Artin Algebra

or level curves in Stewart Calculus

1. So I think I understand correctly when I say that I think that:

Zero sets of polynomial equations (I think Tu should say polynomial functions) over smooth manifolds are closed because by the fundamental theorem of algebra, zero sets are finite, and finite sets are closed because smooth manifolds are Hausdorff.

Is that correct?

2. Can we generalize?

Level sets of polynomial equations (I think Tu should say polynomial functions) over topological manifolds are closed because by the fundamental theorem of algebra, level sets are finite, and finite sets are closed because topological manifolds are Hausdorff.

3. Is this alternative method correct? It is based on vociferous_rutabaga's comment, which I read right after I posted this question:

Polynomial functions are continuous, preimages of closed sets under continuous functions are closed, $\{0\}$ is finite and thus closed by Hausdorff of the manifold of the range.

4. Am I correct to say that:

   • in the preceding proof, I use Hausdorff of the manifold of the range and not Hausdorff of the manifold of the domain and then

   • in the proofs in numbers 1 and 2, I use Hausdorff of the manifold of the domain and not Hausdorff of the manifold of the range

?

Answer 1

This has nothing to do with the fundamental theorem of algebra. The zero set of a polynomial map is closed because

1. Polynomial maps are continuous.
2. The singleton set $\{0\}$ is closed.

So the preimage $f^{-1}(0)$ is closed.

Generalizing, if $F\colon X\to Y$ is any continuous map between topological spaces and $c\in Y$ is a closed point (every point is closed if $Y$ is Hausdorff, or even $T_1$), then the level set $F^{-1}(c)$ is closed.

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Now let me address a few mistakes in your question.

Zero sets of polynomial equations are finite...

This is true for polynomial equations in a single variable. But it's definitely not true for polynomial equations in multiple variables (André 3000 gave a counterexample in the comments). When Tu writes that $\text{SL}(n,\mathbb{R})$ is a zero set of a polynomial equation on $\text{GL}(n,\mathbb{R})$, he's referring to the polynomial equation expressing that the determinant of a matrix is equal to $1$, which is a polynomial in $n^2$ variables.

... by the fundamental theorem of algebra.

The fact that a polynomial equation in one variable has finitely many roots is not the fundamental theorem of algebra. It's much more basic, and it holds over any field (or integral domain). The fundamental theorem of algebra says that every non-constant polynomial in one variable with coefficients in $\mathbb{C}$ has a root in $\mathbb{C}$.

I think Tu should say polynomial functions.

We have a polynomial $\det(A)$ in $n^2$ variables, arranged as the entries of an $(n\times n)$ matrix $A$. Tu wants to think of $\text{SL}(n,\mathbb{R})$ as the zero set of the polynomial equation $\det(A) = 1$, i.e. the set $\{A\in \text{GL}(n,\mathbb{R})\mid \det(A) = 1\}$.

Alternatively, you could look at the the polynomial function $\det\colon \text{GL}(n,\mathbb{R})\to \mathbb{R}$ and define $\text{SL}(n,\mathbb{R})$ as the level set $\det^{-1}(1)$. Or alternatively you could define the polynomial map $F(A) = \det(A) - 1$ and consider the zero set $F^{-1}(0)$. All of these are equivalent.